Vexing cosines

Geometry Level 4

$\cos ^{ 3 }15^\circ+\cos ^{ 3 }25^\circ+\cos ^{ 3 }55^\circ+\cos ^{ 3 }65^\circ+ \cos ^{ 3 }95^\circ+ \\ \cos ^{ 3 }105^\circ+\cos ^{ 3 }135^\circ+\cos ^{ 3 }145^\circ+\cos ^{ 3 }185^\circ$

Find the value of the sum above correct to three decimals.

The answer is 0.00.

3 solutions

Milind Prabhu
Jun 6, 2016

Since, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca),$ .

If $a+b+c=0$ we have $a^3+b^3+c^3=3abc$ .

We have,

$\cos { x } +\cos { ({ 120 }^{ \circ }-x) } +\cos { ({ 120 }^{ \circ }+x)=0 }$ Therefore, ${ (\cos { x) } }^{ 3 }{ +(\cos { { (120 }^{ \circ }-x)) } }^{ 3 }{ +(\cos { { (120 }^{ \circ }+x)) } }^{ 3 }=3\cos { x } \cos { ({ 120 }^{ \circ }-x) } \cos { ({ 120 }^{ \circ }+x) } =\frac { 1 }{ 4 } \cos { 3x }$

(Try to prove it!) Grouping the terms and simplifying we get,

$(\cos ^{ 3 } 15^{ \circ }+\cos ^{ 3 } 135^{ \circ }+\cos ^{ 3 } 105^{ \circ })+(\cos ^{ 3 } 25^{ \circ }+\cos ^{ 3 } 145^{ \circ }+\\ \cos ^{ 3 } 95^{ \circ })+(\cos ^{ 3 } 65^{ \circ }+\cos ^{ 3 } 185^{ \circ }+\cos ^{ 3 } 55^{ \circ })=\frac { 1 }{ 4 } (\cos { 45 } ^{ \circ }+\cos { 75 } ^{ \circ }+\cos { 195 } ^{ \circ })$

Since, $\cos { x } +\cos { ({ 120 }^{ \circ }-x) } +\cos { ({ 120 }^{ \circ }+x)=0 }$ we conclude that the sum is $\frac { 1 }{ 4 } (\cos { 45 } ^{ \circ }+\cos { 75 } ^{ \circ }+\cos { 195 } ^{ \circ })=\frac { 1 }{ 4 }(0)=\boxed { 0 }$

and we are done!

Moderator note:

The terms in the problem seem somewhat arbitrary, and that detracts from the underlying mathematics. Your solution makes sense from the point of view of the problem creator, but it's harder to a problem solver to approach it your way. In this regard, I prefer the other two solutions which start off with $\cos^3 \theta = \frac{1}{4} \cos 3 \theta + \frac{3}{4} \cos \theta$ .